Abstract

In §2 a mapping of nonnegative functions is defined to be an integral if it has the following properties: $I(f) \geqslant 0,I(f) < \infty$ for some $f$, if $f \leqslant g$ then $I(f) \leqslant I(g),I(f + f) = 2I(f),I(\sum \nolimits _{n = 1}^\infty {{g_n}) \leqslant \sum \nolimits _{n = 1}^\infty {I({g_n})} }$. Given an integral $I$ a nonnegative function $f$ is defined to be a measurable function if $I(f + g) = I(f) + I(g)$ for all nonnegative functions $g$. If $f,g,({g_n})_{n = 1}^\infty$ are measurable functions then the following functions are measurable: $f + g,af$ for all $a \geqslant 0,\sum \nolimits _{n = 1}^\infty {{g_n},f - g}$ if $f - g \geqslant 0$ and $I(g) < \infty$; also $\sum \nolimits _{n = 1}^\infty {I({g_n}) = I(\sum \nolimits _{n = 1}^\infty {{g_n})} }$. An example shows if $f,g$ are measurable functions then $\max \{ f,g\}$ may fail to be a measurable function. If an integral has the property that if $f,g$ are measurable functions then $\max \{ f,g\}$ is a measurable function, then the following functions are also measurable: $\min \{ f,g\} ,|f - g|,\sup {g_n}$ and under certain conditions ${\lim _{n \to \infty }}\sup {g_n},\inf {g_n},{\lim _{n \to \infty }}\inf {g_n}$ whenever $({g_n})_{n = 1}^\infty$ is a sequence of measurable functions. A theorem similar to Lebesgue’s dominated convergence theorem is shown to hold. In §1 the Lebesgue integral, which does not in general have the properties required to be an integral as defined in §2, is used to obtain an integral ${\mathbf {U}}$ which does. If $\mu$ is an outer measure and ${\mathfrak {M}_\mu }$ is the $\sigma$-algebra of $\mu$-measurable sets then the set of measurable functions defined in §2 for the integral ${\mathbf {U}}$ contains the usual set of ${\mathfrak {M}_\mu }$-measurable functions. ${\mathbf {U}}$ has the property that if $f$ is a ${\mathfrak {M}_\mu }$-measurable function and if $\int _X {fd\mu }$ denotes the Lebesgue integral of $f$ on a set $X$ then $\int _X {fd\mu = {{\mathbf {U}}_X}fd\mu }$. In §3 it is shown that an integral $I$ defined on a set $X$ induces an outer measure $\mu$. If $\mu$ is a regular outer measure, a representation theorem holds for $I$: if $f$ is a nonnegative function and ${\mathbf {U}}$ is the integral of §1 then $I(f) = {U_X}fd\mu$. Regardless of whether or not the outer measure $\mu$ is regular a similar theorem can be obtained: if $f$ is a nonnegative ${\mathfrak {M}_\mu }$-measurable function then $I(f) = {{\mathbf {U}}_X}fd\mu$. The relationship between $\mu$-measurable sets and measurable functions is explored.

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